jmc

We write $x^2+y^2+z^2=1$ as $x^2+a{y}^2+(1-a)y^2+z^2=1,$ where $a$ is some real number, $0\le a\le 1,$ that is to be determined. Then by AM-GM, $$1=x^2+a{y}^2+(1-a)y^2+z^2\ge 2xy\sqrt{a}+2yz\sqrt{1-a}.$$To make this look more like $\lambda xy+yz,$ we choose $a$ so that the coefficients are proportional, i.e. $$\frac{2\sqrt{a}}{\lambda }=2\sqrt{1-a}.$$Then $\sqrt{a}=\lambda \sqrt{1-a},$ so $a={\lambda }^2(1-a).$ Solving for $a,$ we find $a=\frac{{\lambda }^2}{{\lambda }^2+1}.$

Therefore, $$\frac{2\lambda xy}{\sqrt{{\lambda }^2+1}}+\frac{2yz}{\sqrt{{\lambda }^2+1}}\le 1,$$so $$\lambda xy+yz\le \frac{\sqrt{{\lambda }^2+1}}{2}.$$We want this to equal $\frac{\sqrt{5}}{2},$ so $\lambda =\boxed{2}.$ Equality occurs when $x=\frac{\sqrt{10}}{5},$ $y=\frac{\sqrt{2}}{2},$ and $z=\frac{\sqrt{10}}{10}.$